integral from -7 to 0 (5+ sqrt(49-x^2)) dx
in this question here where did the (1/4)π(7)^2 come from ? im talking about the one thats the best answer..thanks
http://answers.yahoo.com/question/index?qid=20111112160442AAYeyhO
Note that you can split up the integral to get:
∫ [5 + √(49 - x^2)] dx (from x=-7 to 0)
= ∫ 5 dx (from x=-7 to 0) + ∫ √(49 - x^2) dx (from x=-7 to 0).
The first integral represents the area under y = 5 on the interval (-7, 0), which is a rectangle of width 7 and height 5. This rectangle has area (5)(7) = 35, so:
∫ 5 dx (from x=-7 to 0) = 35.
The second integrand is y = √(49 - x^2), which is the upper semi-circle centered at the origin with a radius of 7. The interval (-7, 0) represents the left half of the semi-circle, so ∫ √(49 - x^2) dx (from x=-7 to 0) just represents the area of a quarter circle of radius 7. Thus:
∫ √(49 - x^2) dx = (1/4)π(7)^2 = 49π/4.
Therefore, adding these two results together gives the required value of the integral to be 35 + 49π/4.
I hope this helps!
1 year ago
in this question here where did the (1/4)π(7)^2 come from ? im talking about the one thats the best answer..thanks
http://answers.yahoo.com/question/index?qid=20111112160442AAYeyhO
Note that you can split up the integral to get:
∫ [5 + √(49 - x^2)] dx (from x=-7 to 0)
= ∫ 5 dx (from x=-7 to 0) + ∫ √(49 - x^2) dx (from x=-7 to 0).
The first integral represents the area under y = 5 on the interval (-7, 0), which is a rectangle of width 7 and height 5. This rectangle has area (5)(7) = 35, so:
∫ 5 dx (from x=-7 to 0) = 35.
The second integrand is y = √(49 - x^2), which is the upper semi-circle centered at the origin with a radius of 7. The interval (-7, 0) represents the left half of the semi-circle, so ∫ √(49 - x^2) dx (from x=-7 to 0) just represents the area of a quarter circle of radius 7. Thus:
∫ √(49 - x^2) dx = (1/4)π(7)^2 = 49π/4.
Therefore, adding these two results together gives the required value of the integral to be 35 + 49π/4.
I hope this helps!
1 year ago
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x^2 + y^2 = 49 represents a circle of radius 7 centered at the origin. If we solve this for y, we get:
y^2 = 49 - x^2 ==> y = ±√(49 - x^2).
If we pick the positive sign (y = √(49 - x^2)), we get the upper half of this circle.
Now, since √(49 - x^2) is positive, integrating it over any valid interval (where the function is defined --- in this case [-7, 7]) will represent the area between the x-axis and y = √(49 - x^2) on that interval. So, ∫ √(49 - x^2) dx (from x=-7 to 0) represents the area between the x-axis and y = √(49 - x^2) on the interval [-7, 0], which is a quarter circle of radius 7 (here's a picture of this: http://www.wolframalpha.com/input/?i=%E2… ). The area of the entire circle is π(7)^2 = 49π, so the area of a quarter of it is 49π/4, which is what I got there.
I hope this helps!
y^2 = 49 - x^2 ==> y = ±√(49 - x^2).
If we pick the positive sign (y = √(49 - x^2)), we get the upper half of this circle.
Now, since √(49 - x^2) is positive, integrating it over any valid interval (where the function is defined --- in this case [-7, 7]) will represent the area between the x-axis and y = √(49 - x^2) on that interval. So, ∫ √(49 - x^2) dx (from x=-7 to 0) represents the area between the x-axis and y = √(49 - x^2) on the interval [-7, 0], which is a quarter circle of radius 7 (here's a picture of this: http://www.wolframalpha.com/input/?i=%E2… ). The area of the entire circle is π(7)^2 = 49π, so the area of a quarter of it is 49π/4, which is what I got there.
I hope this helps!
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I agree to the first factor 5 give the definite integral 35 over the interval -7 to 0.
I do not agree to sqrt(49-x^2) is a semi circle. It has a different shape from a circle and the area is a quarter of the whole round of a circle.
I do not agree to sqrt(49-x^2) is a semi circle. It has a different shape from a circle and the area is a quarter of the whole round of a circle.
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Are you from greece? Agapi?